Javascript regular expression to extract characters from mid string with optional end character
I would like to extract characters from mid string with optional end character. If the optional end character is not found, extract until end of string. The first characters are S= and the last optional character is &.
Example #1:
"rilaS=testingabc"
should extract:
"testingabc"
Example #2:
"rilaS=testing123&thistest"
should extract:
"testing123"
This is what I have so far (Javascript):
var Str = "rilaS=testing123&thistest";
var tmpStr = Str.match("S=(.*)[&]{0,1}");
var newStr = tmpStr[1];
alert(newStr);
But it does not detect that the end should be the ampersand (if found). Thank you before hand.
Answer (By ggorlen)
var Str = "rilaS=testing123&thistest";
var tmpStr = Str.match("S=([^&]*)");
var newStr = tmpStr[1];
alert(newStr);
javascript regex
asked Nov 27 '18 at 16:27
Robert SmithRobert Smith
303213
add a comment |
I would like to extract characters from mid string with optional end character. If the optional end character is not found, extract until end of string. The first characters are S= and the last optional character is &.
Example #1:
"rilaS=testingabc"
should extract:
"testingabc"
Example #2:
"rilaS=testing123&thistest"
should extract:
"testing123"
This is what I have so far (Javascript):
var Str = "rilaS=testing123&thistest";
var tmpStr = Str.match("S=(.*)[&]{0,1}");
var newStr = tmpStr[1];
alert(newStr);
But it does not detect that the end should be the ampersand (if found). Thank you before hand.
Answer (By ggorlen)
var Str = "rilaS=testing123&thistest";
var tmpStr = Str.match("S=([^&]*)");
var newStr = tmpStr[1];
alert(newStr);
javascript regex
asked Nov 27 '18 at 16:27
Robert SmithRobert Smith
303213
Str.match("S=(.*)[&]");is not enought ?
– V. Sambor
Nov 27 '18 at 16:37
add a comment |
I would like to extract characters from mid string with optional end character. If the optional end character is not found, extract until end of string. The first characters are S= and the last optional character is &.
Example #1:
"rilaS=testingabc"
should extract:
"testingabc"
Example #2:
"rilaS=testing123&thistest"
should extract:
"testing123"
This is what I have so far (Javascript):
var Str = "rilaS=testing123&thistest";
var tmpStr = Str.match("S=(.*)[&]{0,1}");
var newStr = tmpStr[1];
alert(newStr);
But it does not detect that the end should be the ampersand (if found). Thank you before hand.
Answer (By ggorlen)
var Str = "rilaS=testing123&thistest";
var tmpStr = Str.match("S=([^&]*)");
var newStr = tmpStr[1];
alert(newStr);
javascript regex
asked Nov 27 '18 at 16:27
Robert SmithRobert Smith
303213
I would like to extract characters from mid string with optional end character. If the optional end character is not found, extract until end of string. The first characters are S= and the last optional character is &.
Example #1:
"rilaS=testingabc"
should extract:
"testingabc"
Example #2:
"rilaS=testing123&thistest"
should extract:
"testing123"
This is what I have so far (Javascript):
var Str = "rilaS=testing123&thistest";
var tmpStr = Str.match("S=(.*)[&]{0,1}");
var newStr = tmpStr[1];
alert(newStr);
But it does not detect that the end should be the ampersand (if found). Thank you before hand.
Answer (By ggorlen)
var Str = "rilaS=testing123&thistest";
var tmpStr = Str.match("S=([^&]*)");
var newStr = tmpStr[1];
alert(newStr);
javascript regex
javascript regex
asked Nov 27 '18 at 16:27
Robert SmithRobert Smith
303213
asked Nov 27 '18 at 16:27
Robert SmithRobert Smith
303213
edited Nov 27 '18 at 16:49
Robert Smith
asked Nov 27 '18 at 16:27
Robert SmithRobert Smith
303213
asked Nov 27 '18 at 16:27
Robert SmithRobert Smith
303213
asked Nov 27 '18 at 16:27
Robert SmithRobert Smith
303213
303213
Str.match("S=(.*)[&]");is not enought ?
– V. Sambor
Nov 27 '18 at 16:37
add a comment |
Str.match("S=(.*)[&]");is not enought ?
– V. Sambor
Nov 27 '18 at 16:37
Str.match("S=(.*)[&]"); is not enought ?– V. Sambor
Nov 27 '18 at 16:37
Str.match("S=(.*)[&]"); is not enought ?– V. Sambor
Nov 27 '18 at 16:37
add a comment |
2 Answers
2
active
oldest
votes
You may use /S=([^&]*)/ to grab from an S= to end of line or &:
["rilaS=testingabc", "rilaS=testing123&thistest"].forEach(s =>
console.log(s.match(/S=([^&]*)/)[1])
);answered Nov 27 '18 at 16:34
ggorlenggorlen
7,1883826
1
Thank you worked perfect!
– Robert Smith
Nov 27 '18 at 16:48
add a comment |
Just in case you are wondering why your original regex didn't work: the problem is that the (.*) pattern is greedy - meaning it will happily slurp up anything, including &, and not leave it for for later items to match. This is why you want the "not &" - it will match up to, but not including the &.
answered Nov 27 '18 at 16:57
TroffTroff
101
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You may use /S=([^&]*)/ to grab from an S= to end of line or &:
["rilaS=testingabc", "rilaS=testing123&thistest"].forEach(s =>
console.log(s.match(/S=([^&]*)/)[1])
);answered Nov 27 '18 at 16:34
ggorlenggorlen
7,1883826
1
Thank you worked perfect!
– Robert Smith
Nov 27 '18 at 16:48
add a comment |
You may use /S=([^&]*)/ to grab from an S= to end of line or &:
["rilaS=testingabc", "rilaS=testing123&thistest"].forEach(s =>
console.log(s.match(/S=([^&]*)/)[1])
);answered Nov 27 '18 at 16:34
ggorlenggorlen
7,1883826
1
Thank you worked perfect!
– Robert Smith
Nov 27 '18 at 16:48
add a comment |
You may use /S=([^&]*)/ to grab from an S= to end of line or &:
["rilaS=testingabc", "rilaS=testing123&thistest"].forEach(s =>
console.log(s.match(/S=([^&]*)/)[1])
);answered Nov 27 '18 at 16:34
ggorlenggorlen
7,1883826
You may use /S=([^&]*)/ to grab from an S= to end of line or &:
["rilaS=testingabc", "rilaS=testing123&thistest"].forEach(s =>
console.log(s.match(/S=([^&]*)/)[1])
);["rilaS=testingabc", "rilaS=testing123&thistest"].forEach(s =>
console.log(s.match(/S=([^&]*)/)[1])
);["rilaS=testingabc", "rilaS=testing123&thistest"].forEach(s =>
console.log(s.match(/S=([^&]*)/)[1])
);answered Nov 27 '18 at 16:34
ggorlenggorlen
7,1883826
edited Nov 27 '18 at 16:41
answered Nov 27 '18 at 16:34
ggorlenggorlen
7,1883826
answered Nov 27 '18 at 16:34
ggorlenggorlen
7,1883826
answered Nov 27 '18 at 16:34
ggorlenggorlen
7,1883826
7,1883826
1
Thank you worked perfect!
– Robert Smith
Nov 27 '18 at 16:48
add a comment |
1
Thank you worked perfect!
– Robert Smith
Nov 27 '18 at 16:48
1
1
Thank you worked perfect!
– Robert Smith
Nov 27 '18 at 16:48
Thank you worked perfect!
– Robert Smith
Nov 27 '18 at 16:48
add a comment |
Just in case you are wondering why your original regex didn't work: the problem is that the (.*) pattern is greedy - meaning it will happily slurp up anything, including &, and not leave it for for later items to match. This is why you want the "not &" - it will match up to, but not including the &.
answered Nov 27 '18 at 16:57
TroffTroff
101
add a comment |
Just in case you are wondering why your original regex didn't work: the problem is that the (.*) pattern is greedy - meaning it will happily slurp up anything, including &, and not leave it for for later items to match. This is why you want the "not &" - it will match up to, but not including the &.
answered Nov 27 '18 at 16:57
TroffTroff
101
add a comment |
Just in case you are wondering why your original regex didn't work: the problem is that the (.*) pattern is greedy - meaning it will happily slurp up anything, including &, and not leave it for for later items to match. This is why you want the "not &" - it will match up to, but not including the &.
answered Nov 27 '18 at 16:57
TroffTroff
101
Just in case you are wondering why your original regex didn't work: the problem is that the (.*) pattern is greedy - meaning it will happily slurp up anything, including &, and not leave it for for later items to match. This is why you want the "not &" - it will match up to, but not including the &.
answered Nov 27 '18 at 16:57
TroffTroff
101
answered Nov 27 '18 at 16:57
TroffTroff
101
answered Nov 27 '18 at 16:57
TroffTroff
101
answered Nov 27 '18 at 16:57
TroffTroff
101
101
add a comment |
add a comment |
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Str.match("S=(.*)[&]");is not enought ?– V. Sambor
Nov 27 '18 at 16:37